Question

6.5 practice
answer the questions below.
return to the previous page by closing this tab and clicking on the correct tab above.
do not hit the return button in the top right corner.
1
at a particular high school, students must start their first class at 8:40 am. the times students generally arrive at school are normally distributed with a mean of 8:30 am and a standard deviation of 5 minutes. on a typical day, approximately what percent of the students will be late?
2.3
1.8
4.9
5.4

Explanation:

Step1: Calculate the z - score

The formula for the z - score is \(z=\frac{x - \mu}{\sigma}\), where \(x\) is the value, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
Here, \(x = 8:40\) (which is \(8:30+10\) minutes), \(\mu = 8:30\) (mean), \(\sigma = 5\) (standard deviation).
\(z=\frac{40 - 30}{5}=\frac{10}{5}=2\)

Step2: Use the empirical rule (68 - 95 - 99.7 rule)

For a normal distribution:

• Approximately \(68\%\) of the data lies within \(z=- 1\) and \(z = 1\)
• Approximately \(95\%\) of the data lies within \(z=-2\) and \(z = 2\)
• Approximately \(99.7\%\) of the data lies within \(z=-3\) and \(z = 3\)

The percentage of data beyond \(z = 2\) (in the right - tail) is \(\frac{100\%-95\%}{2}=2.5\%\approx2.3\%\) (using more accurate values from the standard normal table, the value for \(z = 2\) is \(P(Z>2)=1 - P(Z\leq2)\), and from the standard normal table \(P(Z\leq2)=0.9772\), so \(P(Z > 2)=1 - 0.9772=0.0228\approx2.3\%\))

Answer:

2.3